0:06

So good.

Â This is the vector form, the coordinate independent form.

Â Let's now actually apply some particular frames and

Â there's different answers that we get out of this.

Â It's always the same answer, but it's different expressions,

Â different formulations of that same answer.

Â So if I'm using an orbit frame, and here I've used an aircraft-like frame,

Â where if an airplane is flying you have your nose pointing in the one direction,

Â your wings may be to the left or right.

Â I have yaw axis up, aircraft probably would have had it down.

Â But with orbits we tend to have our third axis or

Â one of the axis up in the radial direction.

Â So, 01 is my long track, 02 is my cross track orbit normal, and

Â 03 is my orbit radius direction vector.

Â They're all unit direction vectors.

Â That means Rc by definition is always going to be 7,000 kilometers in that

Â direction, all right?

Â And that direction is defined here by 03.

Â 0:57

So, in the O frame, your position vector on a circular orbit in particular,

Â is just going to be constant.

Â It's just going to be 0, 0, Rc, all right, which is kind of nice.

Â So we can write that.

Â But we've seen the spacecraft part, the inertia tensor,

Â we tend to write in the body frame.

Â 1:18

So while the orbit positions could be written in very simple manners,

Â we tend to need it in the body frame.

Â And now, so if this is your position vector pointing up to your satellite, and

Â you've got a coordinate frame on your satellite, depending on your orientation,

Â you can have an infinity of different variations, and most generally,

Â you're going to have to account for three different vector components, for

Â your Rc inertial position vector, always.

Â So, Rc 1, 2, 3, nothing but

Â B frame components of my position vector, different ways to look at it.

Â So we have that.

Â Now we plug this in, I added these this morning.

Â Everything's in the body frame just to make it explicit.

Â 1:59

So we had Rc, cross I Rc.

Â I've picked a principal coordinate frame here just to make,

Â I can always do that for a rigid body, all right.

Â It exists, it makes your analysis a little bit easier.

Â Now you can do this math, this multiplication is what talking about.

Â You get a little bit of scaling, then the cross product, and

Â this is going to be your answer.

Â So this box, the answer's the same as the earlier one after it's

Â expressed everything now in body frame components.

Â But we can again look at more conditions now and

Â try to figure out when will the gravity gradient torque go to zero?

Â Why do I care about it going to zero?

Â For equilibrious that we're about to study,

Â that's thing if you have these conditions and this is the one disturbance you're

Â modeling, if that disturbance acts on it, you won't be at an equilibrious.

Â This is like gravity gradients we talked about last time.

Â This would be a gravity gradient equilibria.

Â If you disturb it, then,

Â this one weigh more than this point makes it restore itself.

Â So what orientations or what shapes give us that?

Â And as you guys mentioned earlier, one way to make these things go to zero

Â is simply these differences all have to vanish.

Â And that means, as you guys already identified, is that all the principal

Â inertias must be equal which is either a sphere or a cube, those kinds of shapes.

Â 3:19

So, there are particular shapes that can make your attitude go to 0.

Â That's not or so your gravity gradient torque go to 0.

Â And that's important, there was a satellite that we built here called Dandy.

Â Anybody know that one?

Â Heard about it a little bit, right?

Â It was a sphere for reason like that so we could do this stuff, and

Â didn't have any other gravity gradient torques disturbing because I really wanted

Â to measure atmospheric force effects, and

Â I didn't need all the disturbances acting on it very much.

Â So that was really leading to this decision, all right, that gave us zero

Â gravity gradient torques, also how the atmospheric drag effects so

Â you didn't have weird panels sticking out, there were other reasons too.

Â So the other way we can make it go to zero is we can look at these components.

Â Are there particular orientations?

Â Remember, Rc one and two and three,

Â the position vector always points to straight outward but

Â then depending on my attitude, I will have different Rc 1, 2, 3 components.

Â Are there different orientations that will force this LG to go to zero?

Â And the answer I'm showing you is Rc is equal to Rc times bi.

Â 4:39

>> Be more specific, what part of the O frame?

Â >> The radius vector.

Â >> Right, because if you say you're adding a body axis relative to the O frame-

Â >> If they're principal inertia axis?

Â >> Then that implies all my principal inertia axis

Â have to line up with the O frame, and that's not true.

Â 5:01

So your second argument is correct, and

Â it's what we really care about is the Rc part here that we have, right?

Â That's in the O3 direction here, it's basically the orbit to radium to axis.

Â And so that what it says is your position vector on that direction has to

Â be a principal axis of your body.

Â 5:21

And that could be B1, B2, or B3.

Â But it's just locking one of your R axis in.

Â I'm going to use these to have three distinct axis, right?

Â So we have, this is a principal axis, this is a principal axis, and

Â this is a principal axis.

Â It just says that if I line up one of those things as a principal axis.

Â Let's do the simple example, right?

Â Here I have my skinny axis in the vertical direction.

Â That satisfies that condition.

Â 5:49

If I rotate about that axis,

Â this is still in equilibrium, because of the symmetry of the stuff, right?

Â So you can see, that orientation doesn't matter.

Â It just matters that one of the principal axes is lined up.

Â This is another one that we identified,

Â where right now the gravity gradients are even and they cancelled each other.

Â 6:06

But I can rotate about that about the gravity direction.

Â And I still have a zero gravity gradient torque, right?

Â That's what this condition means.

Â So it only locks one of my body axis to be aligned with the orbit direction vector.

Â That's it.

Â But it still leaves an infinity of orientations because you can rotate about

Â that local vertical axis.

Â 6:30

Okay, good, yes?

Â >> [INAUDIBLE]

Â >> Thinking more like J2 and that stuff?

Â >> [INAUDIBLE] >> They're pretty small.

Â J2 is about a thousand times smaller than the regular stuff.

Â So if you care about those accuracies, you could do that and you'd have to account

Â for that then in these developments, that would be actual terms.

Â I don't typically see those because this is already a pretty small torque.

Â 7:00

Many simulations don't even include this, but if you do include it, great,

Â and J2 is much, much smaller.

Â If you're going around asteroids, that's a whole different problem.

Â If you're looking at an asteroid binary system,

Â I know some of you this year might be looking at those.

Â 7:16

That's a full on thing.

Â Then you do these expansions like I've done but

Â on two bodies not just treating the Earth as a big point mass, as a big thing.

Â But for Earth, it is very spherical as far as that goes, right?

Â But yep, no, great question.

Â So, this would be modified.

Â The process is the same, but you might have extra conditions to line it up with

Â the local area, and what does that mean?

Â